Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
const double eps = 0.00001;
const int MOD = 1000000007;
const int INF = 1000055555;
const long long LLINF = 1223372000000000555;
const int N = 1e5 + 1e4;
const int M = 1234;
const int MAGIC = 10;
vector<int> merge(vector<int> a, vector<int> b) {
  vector<int> v;
  int i = 0, j = 0;
  while (i < a.size() || j < b.size()) {
    if (v.size() == MAGIC) break;
    if (i == a.size())
      v.push_back(b[j++]);
    else if (j == b.size())
      v.push_back(a[i++]);
    else if (a[i] < b[j])
      v.push_back(a[i++]);
    else
      v.push_back(b[j++]);
    if (v.size() > 1 && v[v.size() - 2] == v.back()) v.pop_back();
  }
  return v;
}
vector<int> g[N];
vector<int> arr[N][20];
int tin[N], tout[N], timer = 0;
int up[N][20];
void dfs(int v, int p) {
  tin[v] = ++timer;
  up[v][0] = p;
  for (int i = 1; i < 20; i++) {
    up[v][i] = up[up[v][i - 1]][i - 1];
    int UP = up[v][i - 1];
    arr[v][i] = merge(arr[v][i - 1], arr[UP][i - 1]);
  }
  for (int u : g[v]) {
    if (u != p) dfs(u, v);
  }
  tout[v] = ++timer;
}
bool upper(int a, int b) { return (tin[a] <= tin[b] && tout[a] >= tout[b]); }
int LCA(int a, int b) {
  if (upper(a, b)) return a;
  if (upper(b, a)) return b;
  for (int i = 19; i >= 0; i--) {
    if (!upper(up[a][i], b)) a = up[a][i];
  }
  return up[a][0];
}
void out(vector<int> &v, int a) {
  int k = min(a, ((int)v.size()));
  cout << k << ' ';
  for (int i = 0; i < k; i++) {
    cout << v[i] << ' ';
  }
  cout << endl;
}
void solve() {
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int x;
    cin >> x;
    arr[x][0].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    if (arr[i][0].size() > MAGIC) arr[i][0].resize(MAGIC);
  }
  dfs(1, 1);
  while (q--) {
    int u, v, a;
    cin >> u >> v >> a;
    if (v == u) {
      out(arr[v][0], a);
      continue;
    }
    int lca = LCA(u, v);
    if (u == lca) {
      vector<int> vv;
      for (int i = 19; i >= 0; i--) {
        if (!upper(up[v][i], lca) && up[v][i]) {
          vv = merge(vv, arr[v][i]);
          v = up[v][i];
        }
      }
      vv = merge(vv, arr[u][0]);
      vv = merge(vv, arr[v][0]);
      out(vv, a);
      continue;
    }
    if (v == lca) {
      vector<int> vv;
      for (int i = 19; i >= 0; i--) {
        if (!upper(up[u][i], lca)) {
          vv = merge(vv, arr[u][i]);
          u = up[u][i];
        }
      }
      vv = merge(vv, arr[u][0]);
      vv = merge(vv, arr[v][0]);
      out(vv, a);
      continue;
    }
    vector<int> vv = arr[lca][0];
    for (int i = 19; i >= 0; i--) {
      if (!upper(up[v][i], lca)) {
        vv = merge(vv, arr[v][i]);
        v = up[v][i];
      }
    }
    for (int i = 19; i >= 0; i--) {
      if (!upper(up[u][i], lca)) {
        vv = merge(vv, arr[u][i]);
        u = up[u][i];
      }
    }
    vv = merge(vv, arr[u][0]);
    vv = merge(vv, arr[v][0]);
    out(vv, a);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}